Are they any reductionist proofs where an attacker $\mathcal{I}$ for a well established computationally "hard" problem $\mathsf{Π}$ is employing an attacker $\mathcal{A}$ who we assume is able to break a decisional problem $\mathsf{Γ}$ (i.e: semantic security->guess if $b=1$ or $b=0$) in order $\mathcal{I}$ to break $\mathsf{Π}$ and if so to conclude that $\mathsf{Γ}$ is as secure as $\mathsf{Π}$, meaning that if there exists $\mathcal{A}$, we can build $\mathcal{I}$?
Further explanation:
Typically when there is a reduction to a decisional hard problem then in your construction either you model your security with an adversary that breaking the system means that the attacker should compute something or it has to distinguish something (indistinguishability games in order $\mathcal{A}$ to guess $b$ from an encryption of $x_b$). And this attacker is used as a subroutine by the well established problem attacker in order to break his scheme.
Let’s say that the attacker in our model is able to distinguish encryption $x_0$ from encryption $x_1$. This doesn't mean that the attacker is revealing the underlying key, nor that it has to compute something. It means that it can only distinguish. I have hard understandings how such attackers can be used as subroutines for well-established computational problems where an attacker $\mathcal{B}$ has to compute something instead of distinguish (or decide). And I would like to know if such reductions exist or if – from theory – you can never reduct from decisional to computational. I.e: CDH cannot use a DDH attacker to compute $g^{ab}$ but it is the other way around.